3450: Tyvj1952 Easy 概率与期望 DP

维护一个最长连续’o’序列的期望就好了。。

#include<iostream>
#include<cstdio>
using namespace std;
int n;
char s[300005];
double p1[300005],p2[300005];
double ans;
int main()
{
    scanf("%d",&n);
    scanf("%s",s+1);
    for (int i=1;i<=n;i++)
    {
        double p;
        switch (s[i])
        {
            case 'o':p=1.0; break;
            case 'x':p=0.0; break;
            case '?':p=0.5; break;
        }
        p1[i]=(p1[i-1]+1)*p;
        p2[i]=(p2[i-1]+2*p1[i-1]+1)*p;
        ans+=p2[i-1]*(1-p);
    }
    ans+=p2[n];
    printf("%.4lf",ans);
    return 0;
}